From ed9caec5c1e5e2ded244d39dd91afd83758b6a8f Mon Sep 17 00:00:00 2001
From: ANSHUMAAN <75872316+amino19@users.noreply.github.com>
Date: Wed, 2 Jun 2021 17:10:35 +0530
Subject: [PATCH] Update finding_number_of_Digits_in_a_Number.cpp

---
 math/finding_number_of_Digits_in_a_Number.cpp | 27 +++++++++----------
 1 file changed, 12 insertions(+), 15 deletions(-)

diff --git a/math/finding_number_of_Digits_in_a_Number.cpp b/math/finding_number_of_Digits_in_a_Number.cpp
index 67c982cf6..35fa104f8 100644
--- a/math/finding_number_of_Digits_in_a_Number.cpp
+++ b/math/finding_number_of_Digits_in_a_Number.cpp
@@ -1,15 +1,15 @@
 /**
  * @author [ANSHUMAAN](https://github.com/amino19)
- * @file [Program to count digits in an
- * integer](https://www.geeksforgeeks.org/program-count-digits-integer-3-different-methods)
+ * @file
  *
- * @brief Finding number of Digits in a Number
+ * @brief [Program to count digits 
+ * in an integer](https://www.geeksforgeeks.org/program-count-digits-integer-3-different-methods)
  * @details It is a very basic math of finding number of digits in a given
  * number i.e, we can use it by inputting values whether it can be a
- * positive/negative value, lets say integer. There is also second method to do
- * it with, by using "K = floor(log10(N) + 1)", but its only applicable for
+ * positive/negative value, let's say: an integer. There is also a second method: 
+ * by using "K = floor(log10(N) + 1)", but it's only applicable for
  * numbers (not integers).
- * For more details, refer Algorithms-Explanation
+ * For more details, refer to the [Algorithms-Explanation](https://github.com/TheAlgorithms/Algorithms-Explanation/blob/master/en/Basic%20Math/Finding the number of digits in a number.md) repository.
  */
 
 #include <cassert>   /// for assert
@@ -20,7 +20,7 @@
  * @returns 0 on exit
  */
 int main() {
-    // Initialize 'n' & 'count' by 0
+    // Initialize `n` & `count` by 0
     int n = 0;
     int count = 0;
 
@@ -30,18 +30,15 @@ int main() {
     std::cout << "Enter an integer: ";
     std::cin >> n;
 
-    /* iterate until n becomes 0
-     * remove last digit from n in each iteration
-     * increase count by 1 in each iteration */
+    // iterate until `n` becomes 0
+    // remove last digit from `n` in each iteration
+    // increase `count` by 1 in each iteration 
     while (n != 0) {
-        // we can also use: n = n/10
+        // we can also use `n = n / 10`
         n /= 10;
-        /*  each time while loop running, count will
-         *  be increment by 1.
-         */
+        // each time the loop is running, `count` will be incremented by 1.
         ++count;
     }
-
     std::cout << "Number of digits: " << count;
 
     return 0;